Let \(R\) be a ring (not assumed to have an identity) with the property that \(x^2 = x\) for all \(x \in R\) (a Boolean ring).
Prove that \(2x = 0\) for all \(x \in R\) (that is, \(R\) has characteristic 2).
Prove that \(R\) is commutative.
Solution
Goal: Prove that every Boolean ring has characteristic 2 in (a) and is commutative in (b) by evaluating the idempotent property on sums.
1. Part (a): \(2x = 0\) for all \(x \in R\).
1.1. Let \(x \in R\). By hypothesis, \(x + x \in R\) satisfies the idempotent property \((x + x)^2 = x + x\).
1.2. Expand the left-hand side using the distributive laws of ring arithmetic:\[(x + x)^2 = (x + x)(x + x) = x^2 + x^2 + x^2 + x^2.\]
1.3. Since \(x^2 = x\), the expansion simplifies to:\[(x + x)^2 = x + x + x + x = 4x.\]
1.4. Equating the two expressions gives\[4x = 2x.\]
1.5. Subtracting \(2x\) from both sides in the abelian group \((R, +)\) yields\[2x = 0 \quad \text{for all } x \in R.\]
1.6. In particular, \(x = -x\) for all \(x \in R\).
2. Part (b): \(R\) is commutative (\(x y = y x\) for all \(x, y \in R\)).
2.1. Let \(x, y \in R\). By hypothesis, the element \(x + y \in R\) satisfies \((x + y)^2 = x + y\).
2.2. Expand the left-hand side using distributivity:\[(x + y)^2 = x^2 + x y + y x + y^2 = x + x y + y x + y.\]
2.3. Equating this to \(x + y\) gives\[x + x y + y x + y = x + y.\]
2.4. Subtracting \(x + y\) from both sides yields\[x y + y x = 0.\]
2.5. Rearranging gives \(x y = -y x\).
2.6. By Part (a), \(-z = z\) for every \(z \in R\). Applying this to \(z = y x\) gives\[x y = y x.\]
2.7. Since \(x, y \in R\) were arbitrary, \(R\) is commutative.
3. Conclusion:
Every Boolean ring satisfies $2x = 0$ and $x y = y x$.